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3 years ago

Which sequence shows the numbers in order from least to

Mathematics

1 answer:

cestrela7 [59]3 years ago

6 0

Answer:

Step-by-step explanation:

Just did it

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1 2/5m - 3/5(2/3m+1)

erastova [34]

Answer:

step by step

STEP

Equation at the end of step 1

((((3•(m3))+5m2)-5m)+1)

(———————————————————————•m)-1

STEP

Equation at the end of step

(((3m3+5m2)-5m)+1)

(——————————————————•m)-1

STEP

3m3 + 5m2 - 5m + 1

Simplify ——————————————————

Checking for a perfect cube :

3.1 3m3 + 5m2 - 5m + 1 is not a perfect cube

Trying to factor by pulling out :

3.2 Factoring: 3m3 + 5m2 - 5m + 1

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: -5m + 1

Group 2: 5m2 + 3m3

Pull out from each group separately :

Group 1: (-5m + 1) • (1) = (5m - 1) • (-1)

Group 2: (3m + 5) • (m2)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

Polynomial Roots Calculator :

3.3 Find roots (zeroes) of : F(m) = 3m3 + 5m2 - 5m + 1

Polynomial Roots Calculator is a set of methods aimed at finding values of m for which F(m)=0

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers m which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number P/Q then P is a factor of the Trailing Constant and Q is a factor of the Leading Coefficient

In this case, the Leading Coefficient is 3 and the Trailing Constant is 1.

The factor(s) are:

of the Leading Coefficient : 1,3

of the Trailing Constant : 1

Let us test ....

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 8.00

-1 3 -0.33 3.11

1 1 1.00 4.00

1 3 0.33 0.00 3m - 1

The Factor Theorem states that if P/Q is root of a polynomial then this polynomial can be divided by q*x-p Note that q and p originate from P/Q reduced to its lowest terms

In our case this means that

3m3 + 5m2 - 5m + 1

can be divided with 3m - 1

Polynomial Long Division :

3.4 Polynomial Long Division

Dividing : 3m3 + 5m2 - 5m + 1

("Dividend")

By : 3m - 1 ("Divisor")

dividend 3m3 + 5m2 - 5m + 1

- divisor * m2 3m3 - m2

remainder 6m2 - 5m + 1

- divisor * 2m1 6m2 - 2m

remainder - 3m + 1

- divisor * -m0 - 3m + 1

remainder 0

Quotient : m2+2m-1 Remainder: 0

Trying to factor by splitting the middle term

3.5 Factoring m2+2m-1

The first term is, m2 its coefficient is 1 .

The middle term is, +2m its coefficient is 2 .

The last term, "the constant", is -1

Step-1 : Multiply the coefficient of the first term by the constant 1 • -1 = -1

Step-2 : Find two factors of -1 whose sum equals the coefficient of the middle term, which is 2 .

-1 + 1 = 0

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step

(m2 + 2m - 1) • (3m - 1)

(———————————————————————— • m) - 1

STEP

Equation at the end of step 4

m • (m2 + 2m - 1) • (3m - 1)

———————————————————————————— - 1

STEP

Rewriting the whole as an Equivalent Fraction

5.1 Subtracting a whole from a fraction

Rewrite the whole as a fraction using 3 as the denominator :

1 1 • 3

1 = — = —————

1 3

Equivalent fraction : The fraction thus generated looks different but has the same value as the whole

Common denominator : The equivalent fraction and the other fraction involved in the calculation share the same denominator

Adding fractions that have a common denominator :

5.2 Adding up the two equivalent fractions

Add the two equivalent fractions which now have a common denominator

Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible:

m • (m2+2m-1) • (3m-1) - (3) 3m4 + 5m3 - 5m2 + m - 3

———————————————————————————— = ———————————————————————

3 3

Polynomial Roots Calculator :

5.3 Find roots (zeroes) of : F(m) = 3m4 + 5m3 - 5m2 + m - 3

See theory in step 3.3

In this case, the Leading Coefficient is 3 and the Trailing Constant is -3.

The factor(s) are:

of the Leading Coefficient : 1,3

of the Trailing Constant : 1 ,3

Let us test ....

P Q P/Q F(P/Q) Divisor

-1 1 -1.00 -11.00

-1 3 -0.33 -4.04

-3 1 -3.00 57.00

1 1 1.00 1.00

1 3 0.33 -3.00

3 1 3.00 333.00

Polynomial Roots Calculator found no rational roots

Final result :

3m4 + 5m3 - 5m2 + m - 3

———————————————————————

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7 0

3 years ago

Solve for the missing side.<br> 4)<br> 12 mi<br> X<br> 15 mi

Alex_Xolod [135]

The equation is a^2 + b^2 = c^2 and the long slanted side is always c so
12 + b^2 = 15
So using the equation a^2 + b^2 = c^2 and the transverse property the answer is 1.732

5 0

2 years ago

Henry takes out a $650 discounted loan with a simple interest rate of 12% for a period of 7 months. What is the effective intere

Andru [333]

The effective rate of interest will be 9.10 %.

<h3>What is compound interest?</h3>

Compound interest is applicable when there will be a change in principle amount after the given time period.

Let's say you have given 100 for two years with a 10% rate of interest annually than for the second-year principle amount will become 110 instant of 100.

Given for simple interest

Principle amount = $650

Rate of interest = 12%

Time period = 7 months.

Interest= PRT/100

Interest= 650× 12 × 7/100 = 546

So final amount = 650 + 546 = $1196

By compound interest

1196 = 650 $[1 + R/100]^{7}$

R = 9.10%

Hence the effective rate of interest will be 9.10%.

For more information about compound interest,

brainly.com/question/26457073

#SPJ1

6 0

2 years ago

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

3 years ago

Help!!!!!!! True or false

Scrat [10]

The answer to your question is true

3 0

2 years ago